Let f(x) = begin{cases} frac{x^4 - 5x^2 + 4}{ | vedclass

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(D) For $x 
e 1, 2$,the function is $f(x) = \frac{(x^2-1)(x^2-4)}{|(x-1)(x-2)|} = \frac{(x-1)(x+1)(x-2)(x+2)}{|(x-1)(x-2)|}$.Case $x=1$: $\lim_{x 	o

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$R - \{1, 2\}$

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(D) For $x 
e 1, 2$,the function is $f(x) = \frac{(x^2-1)(x^2-4)}{|(x-1)(x-2)|} = \frac{(x-1)(x+1)(x-2)(x+2)}{|(x-1)(x-2)|}$.Case $x=1$: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} \frac{(x-1)(x+1)(x-2)(x+2)}{(x-1)(x-2)} = (1+1)(1+2) = 6$. $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} \frac{(x-1)(x+1)(x-2)(x+2)}{-(x-1)(x-2)} = -(1+1)(1+2) = -6$. Since $\lim_{x 	o 1^+} f(x) 
e \lim_{x 	o 1^-} f(x)$,$f(x)$ is discontinuous at $x=1$.Case $x=2$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} \frac{(x-1)(x+1)(x-2)(x+2)}{(x-1)(x-2)} = (2+1)(2+2) = 12$. $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} \frac{(x-1)(x+1)(x-2)(x+2)}{-(x-1)(x-2)} = -(2+1)(2+2) = -12$. Since $\lim_{x 	o 2^+} f(x) 
e \lim_{x 	o 2^-} f(x)$,$f(x)$ is discontinuous at $x=2$.Thus,$f(x)$ is continuous for all $x \in R - \{1, 2\}$.

Let $f(x) = \begin{cases} \frac{x^4 - 5x^2 + 4}{|(x - 1)(x - 2)|}, & x \ne 1, 2 \\ 6, & x = 1 \\ 12, & x = 2 \end{cases}$. Then $f(x)$ is continuous on the set

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